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Government Polytechnic Muzaffarpur Name of the Lab: Testing & Maintenance of Electrical

Machines Lab

Subject Code: 1620606

Experiment-1

Aim: To perform load test on single phase Induction motor.

Apparatus required:

S No. Apparatus Specification Quantity

1 Voltmeter (0-300V) MI 1

2 Ammeter (0-10A) MI 1

3 Wattmeter (300V, 10A UPF) 1

4 Tachometer (0-10000 RPM) 1

Formula:

Circumference of the brake drum = 2πR (m); where R is the radius of brake drum.

Input power = W watts; where W is the wattmeter reading.

Torque (T) = 9.81*R*(S1-S2) (N-m); where S1, S2 are spring balance readings in Kg.

Output power =

(watts); where N is the speed in RPM.

%efficiency (η) =

Power factor ( ) =

%slip (s) =

;

Where

(rpm)

P = number of poles.

f = supply frequency (Hz).

Precaution:

i. The auto transformer is kept at minimum voltage position.

ii. The motor is started at no load condition.

Procedure:

i. Connections are given as per the circuit diagram

ii. The DPST switch is closed and the single phase supply is given

iii. By adjusting the variac the rated voltage is applied and the corresponding no load

values of speed, spring balance and meter readings are noted down. If the wattmeter

readings show negative deflection on no load, switch of the supply & interchange the

terminals of current coils (M & L) of the wattmeter. Now, again starting the motor

(follow above procedure for starting), take readings.

iv. The procedure is repeated till rated current of the motor.

v. The motor is unloaded, the auto transformer is brought to the minimum voltage

position, and the DPST switch is opened.

vi. The radius of the brake drum is measured.

Observation table:

V

volt

I

amp

Speed

N

(rpm)

Wattmeter

reading

(watts)

Spring balance

readings (Kg)

Torque

T (N-m)

Output

power

(watts)

Power

factor

%effici

ency

(η)

%slip

(s)

S1 S2 S1-S2

OBS ACT

Model graph:

Circuit diagram:

Result:


Machines Lab


Experiment-2

Aim: To conduct load test on given 3-phase squirrel cage induction motor and to plot its

performance characteristics.

Apparatus required:

S No. Apparatus Specification Quantity

1 Voltmeter (0-600V) MI 1

2 Ammeter (0-5A) MI 1

3 Wattmeter (600V, 10A, UPF) 2

4 Tachometer (10000RPM) 1

Formula:

Circumference of brake drum: 2πR (m); where R is the radius.

Input power W =W1+W2 (watts); where W1, W2 are wattmeter reading

Torque (T) = 9.81*R*(S1-S2) (N-m); where S1, S2 are spring balance reading in Kg.

Output power =

(watts)

%efficiency (η) =

Power factor =

√ ; where Power factor

%slip S =

x100

Where Ns = synchronous speed =

(rpm)

P = no. Of poles

F = supply frequency (Hz)

Precaution:

The motor is started at no load condition.

Procedure:

i. Connections are given as per the circuit diagram.

ii. The TPST switch is closed and the 3-phase supply is given.

iii. The motor is started with a DOL starter.

iv. No load readings are noted down.

v. If any one of the wattmeter shows negative deflection, the connections of M and L in

the wattmeter are interchanged after switching off the supply.

vi. Gradually the motor is loaded and in each case all the meter readings are noted down

and the procedure is repeated till the rated current is obtained.

vii. The motor is unloaded, the auto transformer is brought to the minimum voltage

position, and the TPST Switch is opened.

viii. The radius of the brake drum is measured.

Tabular column:

V

volt

I

amp

Speed N

(rpm)

Wattmeter reading

(watts)

Spring balance

reading (Kg)

Torque

(T) N-m

Output

power

(watts)

Power

factor

( )

%effic

iency

(η)

%slip

(s)

W1 W2 W1+

W2

S1 S2 S1-S2

O

bs

A

ct

O

bs

A

ct

Model graph:

Circuit diagram:

Result:


Machines Lab


Experiment-3

To perform polarity test, marking its terminal, selection of appropriate meter and

perform back to back test and compare its regulation with direct loading

Aim: To perform polarity marking on two winding Transformer.

Apparatus required:

S No Apparatus Specification Quantity

1 Two winding transformer 230/115V, 1KVA

2 Voltmeter 0-300V 1


4 Ammeter 0-5A 3

5 Loading rheostat 5KW -

6 Single phase dimmerstat 2KVA -

7 Transformer(teaser with tapping on

primary & secondary)

- -

Circuit diagram:

Procedure:

i. Make the connections as shown in figure.

ii. Connect the primary winding P1 – P2 to supply.

iii. Short circuit the terminals P2 & S2.

iv. Connect the voltmeters across primary & secondary windings of transformer & one

voltmeter across P1 and S1.

v. Switch on the supply.

vi. By varying the input voltage with the help of dimmerstat take various reading V1, V2

and V3 for various steps of input voltage.

vii. Analyse the readings and decide about polarity marking of two windings of

transformer. For this assume that a dot is present at terminal P1 of the primary

winding.

If V3 = (V1 + V2), the transformer has additive polarity and the other dot should be

marked at S2.

If V3 = (V1 – V2), the transformer has subtractive polarity and the other dot should be

marked at S1.

Precaution:

i. All the connection should be perfectly tight.

ii. Supply should not be switched ON until & unless the connections are checked by the

teacher.

iii. Do not bend while taking the readings.

iv. No loose wire should lie on the work-table.

Conclusion:

The given transformer is found to have ____________ polarity. If a dot is marked at P1 on

primary side, the dot on secondary side should be at_________.

Aim: Determination of regulation & efficiency of three-phase Transformer by direct loading.

Apparatus required:


1 Ammeter 0-10A 1

2 Ammeter 0-20A 1



5 Dimmerstat 3-φ AC 10A 1

6 Transformer 440/220V, 5kVA, 3-φ

7 Resistive load

8 Connecting wires

Circuit diagram:

Procedure:

i. Connect the circuit as shown in figure.

ii. Keep load on transformer at off position.

iii. Keeping dimmer stat at zero position, switch on 3-φ supply.

iv. Now increase dimmer stat voltage for 440 V.

v. Note down the no-load readings.

vi. Then increase the load in steps till rated current of the transformer & note down

corresponding readings. Take at least 8 readings.

vii. Calculate efficiency & regulation for each reading.

Observation table:

No-load secondary voltage E2 = ________ volts

S No I1 amp V1 volt W1 watt I2 amp V2 volt W2 watt % reg %η

1

2

3

4

5

Calculation:

x 100

( ) ( )

( ) x 100

Graph:

Plot the graph output power Vs efficiency.

Precaution:

i. All the connections should be perfectly tight.

ii. Supply should be switched ON until and unless the connections are checked by the

teacher.

iii. Do not bend back while taking the reading.

iv. No loose wire should lie on the work table.

v. Thick wire should be used for current circuit and flexible wires for voltage circuits.

vi. The multiplying factor of wattmeter should be correctly noted.

Result:

The % efficiency and regulation of transformer at full load condition is found as follows

% efficiency = ____________%

% regulation = _____________%

Conclusion:

Transformer efficiency initially increases as the load on transformer is increased. After

maximum efficiency if we increase the load further, the efficiency of transformer reduces.

Also terminal voltage reduces as the load is increased.


Machines Lab


Experiment-4

To perform parallel operation of two single phase Transformer

Aim: To operate the given two2KVA, 230/110V single phase Transformers in parallel and

study the load sharing between them when supplying resisters load.

Name plate details:

Single phase transformer

Transformer Transformer I Transformer II

HV side LV side HV side LV side

Rated power

Rated voltage

Rated current

Frequency

Apparatus required:

S No Apparatus Type Range Quantity

1 Ammeter MI

2 Ammeter MI

3 Voltmeter MI

4 Resistive load

5 Single phase variac

6 Single phase transformer

Theory:

It is economical to install numbers of smaller rated transformers in parallel than installing

bigger rated electrical power transformers. This has mainly the following advantages, to

maximize electrical power system efficiency: Generally electrical power transformer gives

the maximum efficiency at full load. If we run numbers of transformers in parallel, we can

switch on only those transformers which will give the total demand by running nearer to its

full load rating for that time. When load increases, we can switch none by one other

transformer connected in parallel to fulfil the total demand. In this way we can run the system

with maximum efficiency.

To maximize electrical power system availability: If numbers of transformers run in parallel,

we can shut down any one of them for maintenance purpose. Other parallel transformers in

system will serve the load without total interruption of power.

To maximize power system reliability: if any one of the transformers run in parallel, is

tripped due to fault of other parallel transformers is the system will share the load, hence

power supply may not be interrupted if the shared loads do not make other transformers over

loaded.

To maximize electrical power system flexibility: There is always a chance of increasing or

decreasing future demand of power system. If it is predicted that power demand will be

increased in future, there must be a provision of connecting transformers in system in parallel

to fulfil the extra demand because, it is not economical from business point of view to install

a bigger rated single transformer by forecasting the increased future demand as it is

unnecessary investment of money. Again if future demand is decreased, transformers running

in parallel can be removed from system to balance the capital investment and its return.

Conditions of parallel operation of transformer:

When two or more transformers run in parallel, they must satisfy the following conditions for

satisfactory performance. These are the conditions for parallel operation of transformers.

i. Same voltage ratio of transformer.

ii. Same percentage impedance.

iii. Same polarity.

iv. Same phase sequence.

Circuit diagram:

Procedure:

i. Make connections as for circuit diagram, keep the load switch and switch S open.

ii. Switch on the mains, see the voltmeter reading of V1, if this reading is 460V (double

the secondary voltage of both the machines) then switch off and interchange the

connections of secondary of any transformer. If reads zero then the switch S can be

closed, this way the polarities can be checked since wrong polarity will short circuit

the transformers if operated in parallel.

iii. Close switch S and then close the load switch.

iv. For various values of load current, record terminal voltage, current in two

secondary’s, power supply by the two transformers and the total power,(do not exceed

0 A for total current)

v. Switch of load and switch of main.

vi. Determine the equivalent reactance and resistance of both transformers referred to HV

winding by SC test.

Calculations:

For a given load current IL at an angle φ the current and power supply by each transformer

can be found out by following formula

( ) * ( )+

( ) * ( )+

If S is the load KVA, then the KVA shared by the transformers can be found out

By SA= (S)X{(ZB)/(ZA+ZB)}

SB=(S)X{(ZA)/(ZA+ZB)}

Check the result obtained with the theoretical calculations.

Results:


Machines Lab


Experiment-6

Aim: To perform no load and block rotor test of three phase slip ring induction motor, circle

diagram and equivalent circuit.

Apparatus required:


1 Voltmeter 0-600V MI, 0-300V MI 2

2 Ammeter 0-10A MI, 0-5A MI 2

3 Wattmeter (300V, 10A, UPF), (600V, 5A, LPF) 2

Theory:

A 3-phase induction motor consists of stator, rotor & other associated parts. In the stator, a 3-

phase winding (provided) are displaced in spaceby120. A 3- phase current is fed to the

winding so that a resultant rotating magnetic flux is generated. The rotor starts rotating due to

the induction effect produced due the relative velocity between the rotor Winding &the

rotating flux.

As a general rule, conversion of electrical energy to mechanical energy takes place in to the

rotating part on electrical motor. In DC motors, electrical power is conduct directly to the

armature, i.e. rotating part through brushes and commutator. Hence, in this sense, a DC motor

can be called as 'conduction motor'.

However, in AC motors, rotor does not receive power by conduction but by induction in

exactly the same way as secondary of a two winding T/F receives its power from the primary.

So, these motors are known as Induction motors. In fact an induction motor can be taken as

rotating T/F, i.e, one in which primary winding is stationary and but the secondary is free.

The starting torque of the Induction motor can be increase by improving its p.f. by adding

external resistance in the rotor circuit from the stator connected rheostat, the rheostat

resistance being progressively cut out as the motor gathers speed. Addition of external

resistance increases the rotor impedance and so reduces the rotor current. At first, the effect

of improved p.f. Predominates the current decreasing effect of impedance. So, starting torque

is increased. At time of starting, external resistance is kept at maximum resistance position

and after a certain time, the effect of increased impedance predominates the effect of

improved p.f and so the torque starts decreasing. By this during running period the rotor

resistance being progressively cut-out as the motor attains its speed. In this way, it is possible

to get good starting torque as well as good running torque.

Procedure:

No load test:


ii. Initially the motor is kept at no load condition.

iii. The TPST switch is closed.

iv. By adjusting the 3Φ auto transformer the machine is brought to rated voltage.

v. The ammeter, voltmeter and wattmeter readings are noted down.

Block rotor test:


ii. Load is applied to prevent the rotor from rotating.

iii. Close the TPST switch.

iv. By adjusting the 3Φ auto transformer rated current is allowed to circulate.

v. The ammeter, voltmeter and wattmeter readings are noted down.

Observation table:

No load test:

V0 volts I0 amps W1 watts W2 watts W0 watts

Block rotor test:

VSC volts ISC amps W1 watts W2 watts WSC watts

Equivalent circuit:

Model graph:

Result:

Construction of circle diagram:

By using the data obtained from the no load test and the blocked rotor test, the circle diagram

can be drawn using the following steps:

i. Take reference phasor V as vertical (Y-axis).

ii. Select suitable current scale such that diameter of circle is 20-30 cm.

iii. From no load test, I0 and φ0 are obtained. Draw vector I0 lagging V by angle φ0. This

is line OA.

iv. Draw horizontal line through extremity of I0 i.e. a parallel to horizontal azis.

v. Draw the current ISN calculated from ISC with the same angle lagging V by angle φSC

from origin O. This is phasor OB.

vi. Join AB. The line AB is called output line.

vii. Draw the perpendicular bisector to AB. Extend it to meet line AD as point C. This is

the centre of the circle.

viii. Draw the circle with C as a centre and radius equal to AC. This meets the horizontal

line drawn from A at B.

ix. Draw the perpendicular from B on the horizontal axis to meet AF line at D and meet

horizontal axis at E.

x. Torque line:

The torque line separates stator and rotor copper losses.

The vertical distance BD represents power input at short circuit i.e. WSN which consist

of core loss, stator and rotor copper losses.

FD = DE = fixed loss

AF α sum of stator and rotor copper losses.

Point G is located as

The line AG is called torque line.

Power scale: As AD represents WSN i.e. power input on short circuit at normal voltage, the

power scale can be obtained as

Power scale =

( ) = W/cm

l (BE) = Distance BE in cm

Location of point E (slip ring induction motor):

Transformation ratio

(

)2

Rotor resistance referred to stator.

Thus point G can be obtained by dividing the line BD in the ratio R2:R1

Location of point D (squirrel cage induction motor):

In a squirrel cage motor, the stator resistance can be measured by conducting resistance test.

i.e. stator copper loss = 3 where ISN is phase value.

Neglecting core loss, WSN = stator copper loss + rotor copper loss

i.e. rotor copper loss =

Dividing line BD in this ratio, the point G can be obtained and AG represents torque line.

To get the torque line, join the points A and G.

xi. To find the full load quantities, draw line BK (= full load output/power scale). Now

draw line PK parallel to output line meeting the circle at point P.

xii. Draw line PT parallel to Y-axis meeting output line at Q, torque line at R, constant

loss line at S and X-axis at T.


Machines Lab


Experiment-7

Aim: To perform load test on DC series motor with mechanical load.

Apparatus required:


1 DC series motor with brake drum

arrangement

3HP, 220V, 12A, 1500 rpm 1

2 Voltmeter 0-300V DC 1

3 Ammeter 0-20A DC 1

4 Tachometer 1

Circuit diagram:

Procedure:

i. Connect the circuit as shown in figure.

ii. Keeping some load on the motor, start it with the help of starter.

iii. At this load, note down the speed and also the forces in spring connected to brake

drum.

iv. Note voltmeter as well as ammeter reading.

v. Increase the mechanical load in steps by tightening the rope and note all the readings

again.

vi. Repeat step 5 till the rated current of motor is reached.

vii. Calculate torque and efficiency of motor.

Precaution:

i. All connections should be perfectly tight and no loose wire should lie on the work

table.

ii. Before switching ON the dc supply, ensure that the starter’s moving arm is at its

maximum resistance position.

iii. Do not switch on or operate the D.C. series motor without load.

iv. Before switching on the DC supply, ensure some water inside the drum for cooling

purpose.

v. Do not switch on the supply, until and unless the connections are checked by the

teacher.

Observation table:

Radius of brake drum: ______cm

S No Vm volts Im amps F1 (Kg) F2 (Kg) N (rpm)

Calculation:

Output torque T = [(F1-F2) x 9.81 x r] N-m where r is radius of brake drum.

Output power P0 = 2πNT/60 watts.

Input power Pi = Vm.Im

x 100

Graph:

Plot speed vs torque and output power vs efficiency.

Conclusion:

At light load the motor speed is high and it reduces fast with rise in load.

government polytechnic muzaffarpur - gpmuz.bih.nic.ingpmuz.bih.nic.in/docs/tmeml.pdfgovernment...

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